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Let $E \subset \mathbb{C}$ be a set with the following property: for any sequence of elements $(e_n)_{n \in \mathbb{N}}$ with $e_n \neq e_m$ for $m \neq n$, $e_n \to 0$. Is $E$ necessarily countable?

(Also convergence is in norm, of course).

It seems like this problem should yield to contrapositive, that is, given an uncountable set, I can always find at least one nonrepeating sequence that does not converge to $0$. My idea is that for some $\epsilon >0$, if $E$ is uncountable, there must exist infinitely many distinct elements in the complement of $B_\epsilon (0)$. Then we can construct a sequence not converging to $0$ from that (so the statement is true). Is there a direct proof for the above?

Andrés E. Caicedo
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Rellek
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    Any non-zero $e\in E$ must be isolated. So you can also ask the question if there are uncountable $E\subseteq \Bbb C$ which are only built from isolated points (and zero). Or the other way around: does any uncountable set $E\subseteq\Bbb C$ has at least two accumulation points? – M. Winter Nov 10 '17 at 14:01

2 Answers2

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Take a countable cover $(B_n)_{n \in \mathbb{N}}$ of $\mathbb{C} \setminus \{0\}$ by compact balls. If any $B_n \cap E$ were infinite, then by Bolzano-Weierstrass, $B_n \cap E$ would contain a converging sequence (which cannot converge to $0$), in contradiction to your assumption. Thus, any $B_n \cap E$ is finite, so we have $E = (\bigcup_{n \in \mathbb{N}} B_n \cap E) \cup (E \cap \{0\})$, i.e.$E$ is countable as a countable union of finite sets.

Arno
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    The cover doesn't have to have balls. Can be any kind of compact sets. I personally like annuli better. – Arthur Nov 10 '17 at 14:01
  • Is this a standard problem? I found it on an old qualifying exam and it stumped me for a while. Thanks for the solution. – Rellek Nov 10 '17 at 14:04
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Yes. $E$ is necessarily countable. And we can give an explicit enumeration.

Proof.

There are only finitely many points of $E$ which are outside of $U_\epsilon(0)=\{x\in E\mid |x|<\epsilon\}$ for any $\epsilon>0$, because otherwise a (non-repeating) sequence can stay outside this neighborhood and not converge to zero.

We can build an enumeration of $E$ in the following way: Enumerate all finitely many points in $E$ which are outside of $U_1(0)$. Then enumerate all the finitely many points outside of $U_{1/2}(0)$, then $U_{1/3}(0)$ and so on. Any point despite zero will be enumerated at some point. And if $0\in E$ then append it to the beginning of your enumeration. $\quad\square$

By the way, here is a proof that any uncountable set has uncountably many accumulation points.

Proof.

Let $E$ be an uncountable set with only countably many accumulation points. We consider $E'$ which we obtain from $E$ by removing all the accumulation points. Note that $E'$ is still uncountable.

Now we consider the closed balls $B_r(0)=\{x\in E'\mid |x|\le r\}$. As a compact (hence sequence compact) set without a limit point, it must contain only finitely many points.

Now we can enumerate all point in $E'$ by first enumerating all the finitely many points in $B_1(0)$, then the finitely many points in $B_2(0)$, then $B_3(0)$ and so on. Hence $E'$ is countable. Contradiction. $\quad\square$

M. Winter
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